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Differentials
Equation 
Thermodynamics is the science of 'small steps', where one explores the behavior of a physical system by making variations on known functions
- The dependence of a function on parameters (e.g., $x$ and $y$) is determined, that is, $f(x, y)$.
- Each of these parameters is varied (e.g., $dx$ and $dy$), and the corresponding slope of the variation is identified.
- The aim is to find the relationship between the slope and the already established relationships within thermodynamics.
Mathematically, this is expressed as :
 $ df = \left(\displaystyle\frac{\partial f }{\partial x }\right)_ y dx + \left(\displaystyle\frac{\partial f }{\partial y }\right)_ x dy $ |
The expression
$D_{x, y}f\equiv\left(\displaystyle\frac{\partial f }{\partial x }\right)_ y$
represents the slope in the x-direction with the other variables held constant (in this case, y). It is read as 'partial derivative of f with respect to x, with y held constant'.
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Definition of the first derivative of the Gibbs potential
Equation
In order to calculate the various parameters, it is necessary to be able to differentiate the Gibbs potential, which corresponds to the slopes of this function with respect to pressure or temperature.
In general, the Gibbs potential factors, denoted as $g_x$, are defined with $x$ representing the variable and $g$ representing the molar Gibbs free energy, as follows:
| $ g_x =\displaystyle\frac{\partial g }{\partial x }$ |
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Second derivative of the Gibbs potential
Equation
For the calculation of various parameters, it is necessary to be able to take second-order derivatives of the Gibbs potential, which corresponds to the curvatures of this function with respect to pressure and/or temperature.
In general, the factors of the Gibbs potential are defined as follows:
| $ g_{xy} =\displaystyle\frac{\partial^2 g }{\partial x \partial y }$ |
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Example of application of the method
Description
If we consider the internal energy $U(V,S)$, it depends on two variables:
• The volume $V$
• The entropy $S$
Therefore, its variation can be expressed using the relationship:
| $ df = \left(\displaystyle\frac{\partial f }{\partial x }\right)_ y dx + \left(\displaystyle\frac{\partial f }{\partial y }\right)_ x dy $ |
in the form:
$dU = \left(\displaystyle\frac{\partial U }{\partial V }\right)_ S dV + \left(\displaystyle\frac{\partial U }{\partial S }\right)_ V dS$
According to the first law of thermodynamics, we know that the variation of internal energy $dU$ is equal to:
From this, we can conclude that the slopes are the pressure $p$:
$\left(\displaystyle\frac{\partial U }{\partial V }\right)_ S = -p$
and the temperature $T$:
$\left(\displaystyle\frac{\partial U }{\partial S }\right)_ V = T$
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Ejemplo de potencial termodinámico
Description
To establish the relationships, thermodynamic potentials are introduced, which are potential energies that include or exclude certain forms of energy in a system, such as the energy associated with work $pV$ and the energy associated with entropy $TS$, which cannot be used to perform work.
In the case of enthalpy $H$, it corresponds to the internal energy of the system, which includes the movement of particles, but also incorporates the energy required to form the system, i.e., the work $pV$ done to establish it. Therefore, it is defined as:
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Molar enthalpy
Equation
In addition to the thermodynamic potential itself, its molar version can be defined by simply dividing its magnitude by the molar mass. In the case of enthalpy $H$, this is defined as
| $ h = \displaystyle\frac{ H }{ M_{mol} }$ |
where $M_m$ is the molar mass.
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Molar enthalpy differentiation
Equation
The enthalpy depends on the pressure $p$, entropy $h$, and in our case, also on the salt concentration $i$. Therefore, the respective differences $dh$, $dp$, and $di$ must satisfy:
| $ dh = \left(\displaystyle\frac{\partial h }{\partial s }\right)_{ p , i } ds + \left(\displaystyle\frac{\partial h }{\partial p }\right)_{ p , i } dp + \left(\displaystyle\frac{\partial h }{\partial i }\right)_{ p , s } di $ |
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Molar enthalpy change
Equation
It has been determined that the molar enthalpy $h$ varies as a function of molar entropy $s$, pressure $p$, and salinity $i$ as follows:
| $ dh =T ds + \alpha dp + \mu di$ |
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Slope of enthalpy in entropy
Equation
The slope of the molar enthalpy $h$ with respect to entropy is equal to the temperature $T$:
| $ T = \displaystyle\frac{\partial h }{\partial S }$ |
If we compare the differentiation of enthalpy
| $ dh = \left(\displaystyle\frac{\partial h }{\partial s }\right)_{ p , i } ds + \left(\displaystyle\frac{\partial h }{\partial p }\right)_{ p , i } dp + \left(\displaystyle\frac{\partial h }{\partial i }\right)_{ p , s } di $ |
with its variation
| $ dh =T ds + \alpha dp + \mu di$ |
we can conclude that
| $ T = \displaystyle\frac{\partial h }{\partial S }$ |
ID:(12394, 0)
Pendiente de la presión
Equation
The slope of the molar enthalpy $h$ with respect to entropy is equal to the pressure $p$:
| $ \alpha = \displaystyle\frac{\partial h }{\partial p } $ |
If we compare the differentiation of the enthalpy
| $ dh = \left(\displaystyle\frac{\partial h }{\partial s }\right)_{ p , i } ds + \left(\displaystyle\frac{\partial h }{\partial p }\right)_{ p , i } dp + \left(\displaystyle\frac{\partial h }{\partial i }\right)_{ p , s } di $ |
with its variation
| $ dh =T ds + \alpha dp + \mu di$ |
we can conclude that
| $ \alpha = \displaystyle\frac{\partial h }{\partial p } $ |
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Slope of enthalpy in salinity
Equation
The slope of the molar enthalpy $h$ with respect to entropy is equal to salinity $s$:
| $ \mu = \displaystyle\frac{\partial h }{\partial i }$ |
If we compare the differentiation of the enthalpy
| $ dh = \left(\displaystyle\frac{\partial h }{\partial s }\right)_{ p , i } ds + \left(\displaystyle\frac{\partial h }{\partial p }\right)_{ p , i } dp + \left(\displaystyle\frac{\partial h }{\partial i }\right)_{ p , s } di $ |
with its variation
| $ dh =T ds + \alpha dp + \mu di$ |
we can conclude that
| $ \mu = \displaystyle\frac{\partial h }{\partial i }$ |
.
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Adiabatic lapse rate
Equation
The stability of seawater is characterized by the so-called adiabatic lapse rate, which is directly related to the problem of temperature and salinity gradients that can destabilize the marine water column.
The adiabatic lapse rate is defined as:
| $ \Gamma \equiv \displaystyle\frac{\partial T }{\partial p }$ |
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Adiabatic lapse rate and enthalpy
Equation
The adiabatic lapse rate can be calculated using the effective volume $\alpha$ and the specific heat at constant pressure $c_p$ as follows:
| $ \Gamma = \displaystyle\frac{ \alpha }{ c_p }$ |
The adiabatic lapse rate, given by
| $ \Gamma \equiv \displaystyle\frac{\partial T }{\partial p }$ |
can be expressed in terms of enthalpy using the relationship
| $ T = \displaystyle\frac{\partial h }{\partial S }$ |
and the relationship
| $ \alpha = \displaystyle\frac{\partial h }{\partial p } $ |
as
$\Gamma =\displaystyle\frac{\partial T}{\partial p}=\displaystyle\frac{\partial}{\partial p}\displaystyle\frac{\partial h}{\partial s}=\displaystyle\frac{\partial}{\partial s}\displaystyle\frac{\partial h}{\partial p}=\displaystyle\frac{\partial \alpha}{\partial s}$
therefore, the adiabatic lapse rate is
| $ \Gamma = \displaystyle\frac{ \alpha }{ c_p }$ |
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Variation of entropy with temperature
Equation
The molar entropy varies with temperature according to the following relationship:
| $ \left(\displaystyle\frac{\partial s }{\partial T }\right)_{ p , i } = \displaystyle\frac{ c_p }{ T }$ |
Since the heat capacity at constant pressure is defined through enthalpy as
$c_p=\left(\displaystyle\frac{\partial h }{\partial T }\right)_{ p , i }$
we have
| $ T = \displaystyle\frac{\partial h }{\partial S }$ |
which implies
$c_p=\displaystyle\frac{\partial h}{\partial T}=\displaystyle\frac{\partial h}{\partial s}\displaystyle\frac{\partial s}{\partial T}=T\displaystyle\frac{\partial s}{\partial T}$
thus, we have the relationship
| $ \left(\displaystyle\frac{\partial s }{\partial T }\right)_{ p , i } = \displaystyle\frac{ c_p }{ T }$ |
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Adiabatic lapse rate without entropy
Equation
The adiabatic lapse rate can be calculated using the equation:
| $ \Gamma = \displaystyle\frac{ T }{ c_p } \displaystyle\frac{ \partial \alpha }{ \partial T } $ |
With the adiabatic lapse rate given by
| $ \Gamma = \displaystyle\frac{ \alpha }{ c_p }$ |
we have
| $ \left(\displaystyle\frac{\partial s }{\partial T }\right)_{ p , i } = \displaystyle\frac{ c_p }{ T }$ |
that the adiabatic lapse rate can be written as
$\Gamma=\displaystyle\frac{\partial \alpha }{\partial s }=\displaystyle\frac{\partial \alpha }{\partial T }\displaystyle\frac{\partial T }{\partial s }=\displaystyle\frac{ T }{ c_p }\displaystyle\frac{\partial \alpha }{\partial T }$
we have
| $ \Gamma = \displaystyle\frac{ T }{ c_p } \displaystyle\frac{ \partial \alpha }{ \partial T } $ |
where $T$ is the temperature, $c_p$ is the specific heat capacity at constant pressure, and $\partial\alpha/\partial T$ is the variation of the relative volume with respect to temperature.
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Adiabatic lapse rate and properties
Equation
The adiabatic lapse rate can be calculated using the temperature $T$, the specific heat capacity at constant pressure $c_p$, the thermal expansion coefficient $k_T$, and the density $\rho$, as follows:
| $ \Gamma = \displaystyle\frac{ T }{ c_p }\displaystyle\frac{ k_T }{ \rho }$ |
With the definition of the specific volume
| $ \alpha = \displaystyle\frac{1}{ \rho }$ |
and the relationship for thermal expansion given by
| $ k_T =\displaystyle\frac{1}{ \alpha }\left(\displaystyle\frac{ \partial\alpha }{ \partial T }\right)_{ p , S }$ |
the derivative of the specific volume with respect to the adiabatic lapse rate, expressed as
| $ \Gamma = \displaystyle\frac{ T }{ c_p } \displaystyle\frac{ \partial \alpha }{ \partial T } $ |
can be expressed as
| $ \Gamma = \displaystyle\frac{ T }{ c_p }\displaystyle\frac{ k_T }{ \rho }$ |
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